Let `S_n` denote the sum of first n terms of an AP and `3S_n=S_(2n)` What is `S_(3n):S_n` equal to?

Options

A. 4
B. 6
C. 8
D. 10

Explanation

To solve the problem, we need to find the ratio \( S_{3n} : S_n \) given that \( 3S_n = S_{2n} \).

### Step 1: Write the formula for the sum of the first n terms of an AP
The sum of the first \( n \) terms of an arithmetic progression (AP) can be expressed as:
\[ S_n = \frac{n}{2} \left( 2A + (n-1)D \right) \]
where \( A \) is the first term and \( D \) is the common difference.

### Step 2: Write the formula for \( S_{2n} \)
Using the same formula, the sum of the first \( 2n \) terms is:
\[ S_{2n} = \frac{2n}{2} \left( 2A + (2n-1)D \right) = n \left( 2A + (2n-1)D \right) \]

### Step 3: Use the given condition \( 3S_n = S_{2n} \)
Substituting the expressions for \( S_n \) and \( S_{2n} \) into the equation:
\[ 3 \left( \frac{n}{2} \left( 2A + (n-1)D \right) \right) = n \left( 2A + (2n-1)D \right) \]
This simplifies to:
\[ \frac{3n}{2} \left( 2A + (n-1)D \right) = n \left( 2A + (2n-1)D \right) \]

### Step 4: Cancel \( n \) from both sides (assuming \( n \neq 0 \))
Dividing both sides by \( n \):
\[ \frac{3}{2} \left( 2A + (n-1)D \right) = 2A + (2n-1)D \]

### Step 5: Clear the fraction by multiplying by 2
Multiplying through by 2 gives:
\[ 3 \left( 2A + (n-1)D \right) = 4A + (4n-2)D \]

### Step 6: Expand and rearrange the equation
Expanding both sides:
\[ 6A + 3(n-1)D = 4A + (4n-2)D \]
Rearranging gives:
\[ 6A - 4A = (4n-2)D - 3(n-1)D \]
This simplifies to:
\[ 2A = (4n - 2 - 3n + 3)D \]
Thus:
\[ 2A = (n + 1)D \]

### Step 7: Find \( S_{3n} \)
Now, we need to find \( S_{3n} \):
\[ S_{3n} = \frac{3n}{2} \left( 2A + (3n-1)D \right) \]

### Step 8: Substitute \( D = \frac{2A}{n+1} \)
From the earlier step, we have \( D = \frac{2A}{n+1} \). Substitute this into \( S_{3n} \):
\[ S_{3n} = \frac{3n}{2} \left( 2A + (3n-1) \frac{2A}{n+1} \right) \]
This simplifies to:
\[ S_{3n} = \frac{3n}{2} \left( 2A + \frac{(3n-1)2A}{n+1} \right) \]

### Step 9: Find the ratio \( S_{3n} : S_n \)
Now we can find the ratio:
\[ \frac{S_{3n}}{S_n} = \frac{\frac{3n}{2} \left( 2A + \frac{(3n-1)2A}{n+1} \right)}{\frac{n}{2} \left( 2A + (n-1)D \right)} \]
Substituting \( D \) again:
\[ = \frac{3 \left( 2A + \frac{(3n-1)2A}{n+1} \right)}{2A + (n-1) \frac{2A}{n+1}} \]

### Step 10: Simplify the expression
After simplification, we find that:
\[ \frac{S_{3n}}{S_n} = 6 \]

### Final Answer
Thus, the ratio \( S_{3n} : S_n \) is:
\[ \boxed{6} \]